Convergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$

Hint We can apply a cubic analogue of the technique usually called multiplying by the conjugate, which itself is probably familiar from working with limits involving square roots. In this case, write $$u = \sqrt[3]{n^3 + 1},$$ so that $$u^3 - n^3 = (n^3 + 1) - (n^3) = 1.$$ On the other hand, factoring gives (in analogy to the difference of perfect squares factorization $a^2 - b^2 = (a - b)(a + b)$ used in the square root case case) $$u^3 - n^3 = (u - n) (u^2 + un + n^2).$$

Explicitly, multiplying gives that we can write our sum as \begin{align} \sum_{n = 0}^{\infty} (u - n) &= \sum_{n = 0}^{\infty} (u - n) \cdot \frac{u^2 + un + n^2}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{u^3 - n^3}{u^2 + un + n^2} \\ &= \sum_{n = 0}^{\infty} \frac{1}{u^2 + un + n^2} . \end{align} Now, $u > n$, and so the summand satisfies $$\frac{1}{u^2 + un + n^2} < \frac{1}{3 n^2}.$$ The given series is bounded by $$\frac{1}{3} \sum_{n = 0}^{\infty} \frac{1}{n^2},$$ which converges by, e.g., the $p$-test.

For any $x\in[0,1]$, the inequality: $$ (1+x)^{\frac{1}{3}}\leq 1+\frac{x}{3} \tag{1}$$ holds by concavity or just by considering the cube of both terms. It follows that: $$ 0\leq \sqrt[3]{1+n^3}-n = n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\leq\frac{1}{3n^2}\tag{2}$$ hence the given series is convergent by the p-test or by the inequality: $$ 0\leq \sum_{n=1}^{N}\frac{1}{n^2} \leq 1+\sum_{n=1}^{N-1}\frac{1}{n(n+1)}=2-\frac{1}{N}\leq 2.\tag{3}$$

Hint.

Note

$\sqrt[3]{n^3+1} = n\sqrt[3]{1+1/n^3}$, and then $\sqrt[3]{1+1/n^3} = 1 + \frac{1}{3n^3} + o(\frac{1}{n^3})$.

Next, I belive, limit comparison test will work.